# Trouble Understanding Proof Of Invariant Relationship

In part of a proof I am reading this is stated:

$2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) + (a_n + c_n )^2 + (b_n + d_n )^2 ≥ 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ).$ (1)

From this invariant inequality relationship we conclude that, for $n ≥ 2$

$a_n^2 + b_n^2 + c_n^2 + d_n^2 ≥ 2 ^{n - 1} (a_1^2 + b_1^2 + c_1^2 + d_1^2 ).$ (2)

I do not understand how (2) was derived from (1)?

Full proof: Proof

• Should some of the subscript $n$'s in (1) be $1$'s? We don't have the definitions for $a_n, b_n, c_n, d_n$ to see how they depend on $n$. – John Dec 18 '14 at 20:20

They made the substitution $(a_n + c_n )^2 + (b_n + d_n )^2 = -2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$

in the equality:

$$(a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2) \\= 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) - 2a_nb_n - 2b_nc_n - 2c_nd_n - 2d_na_n$$

to write it as:

$$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 = 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) + (a_n + c_n )^2 + (b_n + d_n )^2$$

Hence, $$\displaystyle a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) \text{ for each n \ge 1}$$

(since, $(a_n + c_n )^2 + (b_n + d_n )^2 \ge 0$)

inductively leads to: $$a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2)$$

$$2(a_n^2 + b_n^2 + c_n^2 + d_n^2) \ge 4(a_{n-1}^2 + b_{n-1}^2 + c_{n-1}^2 + d_{n-1}^2)$$

$$\cdots$$

$$2^{n-1} (a_2^2 + b_2^2 + c_2^2 + d_2^2) \ge 2^{n} (a_1^2 + b_1^2 + c_1^2 + d_1^2)$$

Now add these $n$ inequalities !

• I didn't really understand the last part where you from 2(...) to 4 (...) to 2^n, how do you know the coefficient doubles? – fYre Dec 18 '14 at 21:06
• @fYre Its a consequence of using $\displaystyle a_{n+1}^2 + b_{n+1}^2 + c_{n+1}^2 + d_{n+1}^2 \ge 2(a_n^2 + b_n^2 + c_n^2 + d_n^2 ) \text{ for each$n \ge 1$}$ repeatedly ! (each time you replace $n \to n-1$ the coeffn doubles) – r9m Dec 18 '14 at 21:09
• just wondering is there a specific name for the last inductive relations you showed? Like recurrence relation or something similar that I can read about? – fYre Dec 18 '14 at 21:23
• @fYre You could can call it a kind of Recurrence Inequality if you want to ! :-) (my favorite is Ramsey Recurrence Inequality .. just providing an example) – r9m Dec 18 '14 at 21:26